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Simple combinatorial problem

  1. Mar 13, 2010 #1

    quasar987

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    I am trying to compute the number

    [tex]\sum_{g\in S_n}|\mathrm{fix}(g^2)|^2=\sharp\left(\bigsqcup_{g\in S_n}\mathrm{fix}(g^2)\times \mathrm{fix}(g^2)\right)[/tex]

    where S_n is the permutation group on the n elements {1,...n} and where fix(g) = {i in {1,...,n} : g(i)=i}.

    So the plan is to fix a pair (i,j) in {1,...n} x {1,...n} and ask how many g's are such that (i,j) belongs to fix(g²) x fix(g²). Then summing over all the possible values of (i,j) gives the desired answer.

    I start with (i,j) in {1,...n} x {1,...n} such that [itex]i\neq j[/itex]. How many g's are such that (i,j) belongs to fix(g²) x fix(g²)?

    - There are those g's such that g(i)=i, g(j)=j and there are (n-2)! of them.

    - There are those g's such that g(i)=k, g(k)=i ([itex]k\neq i[/itex]), g(j)=j and there are (n-2)(n-3)!=(n-2)! of them.

    - There are those g's such that g(i)=i, g(j)=l, g(l)=j ([itex]l\neq j[/itex]) and there are (n-2)(n-3)!=(n-2)! of them.

    - There are those g's such that g(i)=j, g(j)=i, and there are (n-2)! of them.

    - There are those g's such that g(i)=k, g(k)=i ([itex]k\neq i,j[/itex]), g(j)=l, g(l)=g(j) ([itex]l\neq i,j[/itex]), and there are (n-2)(n-3)(n-4)!=(n-2)! of them.

    So in total, for (i,j), [itex]i\neq j[/itex] there are 5(n-2)! g's in S_n such that (i,j) is in fix(g²) x fix(g²).

    Next, if i in {1,...n} is fixed, how many g's are such that (i,i) is in fix(g²) x fix(g²)?

    - There are those g's such that g(i)=i and there are (n-1)! of them.

    - There are those g's such that g(i)=k, g(k)=i ([itex]k\neq i[/itex]) and there are (n-1)(n-2)!=(n-2)! of them.

    So in total, for each i in {1,...,n}, there are 2(n-1)! g's in S_n such that (i,i) is in fix(g²) x fix(g²).

    Since there are n(n-1) ways of choosing (i,j) such that [itex]i\neq j[/itex] and n ways to choosing i, I conclude that

    [tex]\sum_{g\in S_n}|\mathrm{fix}(g^2)|^2=\sharp\left(\bigsqcup_{g\in S_n}\mathrm{fix}(g^2)\times \mathrm{fix}(g^2)\right) = n(n-1)(5(n-2)!)+n(2(n-1)!)=5n!+2n!=7n![/tex]

    I have reasons to believe that the correct answer is 8n!. Can anyone see which case I've missed?
     
    Last edited: Mar 13, 2010
  2. jcsd
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